A high diver leaves the end of a 6.0 m high diving board and strikes the water 1

Question

A high diver leaves the end of a 6.0 m high diving board and strikes the water 1.4 s later, 4.0 mbeyond the end of the board. Considering the diver as a particle,

Determine the magnitude of her initial velocity, v 0.

Express your answer to two significant figures and include the appropriate units.

Determine the angle of her initial velocity, v 0.

Express your answer to two significant figures and include the appropriate units.

Determine the maximum height reached.

Express your answer to two significant figures and include the appropriate units.

Determine the magnitude of the velocity v f with which she enters the water.

Express your answer to two significant figures and include the appropriate units.

Determine the angle of the velocity v f with which she enters the water.

Express your answer to two significant figures and include the appropriate units.

Explanation / Answer

in horizontal,

(v0 cos(theta)) t = 4

v0 cos(theta) = 4/1.4 = 2.857 ..... (i)

in vertical,

yf - yi = v0y t + ay t^2 /2

0 - 6 = (v0 sin(theta))(1.4) + (-9.8 x 1.4^2 / 2)

v0 sin(theta) = 3.604 .... (ii)

(ii)/(i) => tan(theta) = 3.604 / 2.857

theta = 51.6 deg ....Ans(Angle of initial velocity)

v0 = 4.60 m/s ......Ans (magnitude of initial velocity )

Hmax = 3.604^2 / (2 x 9.8) = 0.66 m from board  

and 0.66 + 6 = 6.66 m from water level

vx = v0x = 2.857 m/s

vy = (3.604) + (-9.8x 1.4) = -10.116 m/s

magnitude of vf= sqrt(vx^2 + vy^2) = 10.5 m/s ....Ans

angle = tan^-1(-10.116/2.857) = -74.2 deg

Or 74.2 below the horizontal

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